It is well-known as Kato's theorem [1]} that if \(V\)  is bounded and \(V=o(|x|^{-1})\)  as \(|x|\rightarrow \infty \)  then the Schrödinger operator \(-\Delta +V\)  has no positive eigenvalues (also see [2]}, [3]}, [4]} for more related results and references). By contraries, such a criterion cannot hold for the fourth-order Schrödinger operator \(H=\Delta ^2+V\) , so the assumption on the absence of positive eigenvalues seems to be indispensable. Indeed, it is easy to construct a Schwartz function \(V(x)\)  so that \(H\)  on \({\mathbb {R}}\)  has an eigenvalue 1.In fact, \(V(x)=20/\cosh ^2(x)-24/\cosh ^4(x)\in {\mathcal {S}}({\mathbb {R}})\)  satisfies \(\frac{d^4\psi _0}{dx^4}+ V(x)\psi _0=\psi _0\)  where \(\psi _0=1/\cosh (x)=2/(e^x+e^{-x})\in L^2(\mathbb {R})\) . Moreover, in any dimensions \(n\ge 1\) , one can also easily construct \(V\in C_0^\infty ({\mathbb {R}}^n)\)  so that \(H\)  has positive eigenvalues (see Feng et al.[5]}). These results clearly indicate that the absence of positive eigenvalues for the fourth-order Schrödinger operator would be more subtle and unstable than the second order cases under the potential perturbation \(V\) .
